Math Doubts

Evaluate $\displaystyle \Large \lim_{x \,\to\, \pi} \, \normalsize \dfrac{1-\cos{7(x-\pi)}}{5{(x-\pi)}^2}$

$x$ is a literal number but considered it to denote the angle of a right angled triangle. The quotient of subtraction of $\cos{7(x-\pi)}$ from one by the product of $5$ and ${(x-\pi)}^2$ is expressed in mathematical form as follows.

$\dfrac{1-\cos{7(x-\pi)}}{5{(x-\pi)}^2}$

The limit of the function is written in the following form as $x$ tends to pi.

$\displaystyle \Large \lim_{x \,\to\, \pi} \, \normalsize \dfrac{1-\cos{7(x-\pi)}}{5{(x-\pi)}^2}$

01

Check the function

Substitute $x$ is equal to $\pi$ to study the functionality of the algebraic trigonometric function as $x$ tends to $\pi$.

$=\,\,\,$ $\dfrac{1-\cos{7(\pi-\pi)}}{5{(\pi-\pi)}^2}$

$=\,\,\,$ $\dfrac{1-\cos{7(0)}}{5{(0)}^2}$

$=\,\,\,$ $\dfrac{1-\cos{0}}{5 \times 0}$

$=\,\,\,$ $\dfrac{1-1}{0}$

$=\,\,\,$ $\dfrac{0}{0}$

The value of the function is indeterminate when the value of the $x$ approaches $\pi$.

02

Transforming the function

Take $u = x-\pi$.

If $x \to \pi$, then $x-\pi \to 0$. Therefore, $u \to 0$. Now, transform the entire mathematical expression in terms of $u$ purely.

$=\,\,\,$ $\displaystyle \Large \lim_{u \,\to\, 0} \, \normalsize \dfrac{1-\cos{7u}}{5u^2}$

03

Simplifying the function

Transform the numerator to apply subtraction trigonometric rule of subtraction of cos of double angle from one.

$=\,\,\,$ $\displaystyle \Large \lim_{u \,\to\, 0} \, \normalsize \dfrac{1-\cos{\Bigg(\dfrac{2}{2} \times 7u\Bigg)}}{5u^2}$

$=\,\,\,$ $\displaystyle \Large \lim_{u \,\to\, 0} \, \normalsize \dfrac{1-\cos{2 \Bigg(\dfrac{7u}{2}\Bigg)}}{5u^2}$

$=\,\,\,$ $\displaystyle \Large \lim_{u \,\to\, 0} \, \normalsize \dfrac{2\sin^2 {\Bigg(\dfrac{7u}{2}\Bigg)}}{5u^2}$

$=\,\,\,$ $\displaystyle \Large \lim_{u \,\to\, 0} \, \normalsize \dfrac{2}{5} \times \dfrac{\sin^2 {\Bigg(\dfrac{7u}{2}\Bigg)}}{u^2}$

Make some adjustments to denominator to be appeared same as the angle of the sine function in numerator.

$=\,\,\,$ $\displaystyle \dfrac{2}{5} \times \Large \lim_{u \,\to\, 0} \, \normalsize \dfrac{\sin^2 {\Bigg(\dfrac{7u}{2}\Bigg)}}{{\Bigg(u \times \dfrac{7}{2} \times \dfrac{2}{7}\Bigg)}^2}$

$=\,\,\,$ $\displaystyle \dfrac{2}{5} \times \Large \lim_{u \,\to\, 0} \, \normalsize \dfrac{\sin^2 {\Bigg(\dfrac{7u}{2}\Bigg)}}{{\Bigg(\dfrac{7u}{2} \times \dfrac{2}{7}\Bigg)}^2}$

Use the product rule of exponents to expand the denominator.

$=\,\,\,$ $\displaystyle \dfrac{2}{5} \times \Large \lim_{u \,\to\, 0} \, \normalsize \dfrac{\sin^2 {\Bigg(\dfrac{7u}{2}\Bigg)}}{{\Bigg(\dfrac{7u}{2}\Bigg)}^2 \times {\Bigg(\dfrac{2}{7}\Bigg)}^2}$

$=\,\,\,$ $\displaystyle \dfrac{2}{5} \times \Large \lim_{u \,\to\, 0} \, \normalsize \dfrac{\sin^2 {\Bigg(\dfrac{7u}{2}\Bigg)}}{{\Bigg(\dfrac{7u}{2}\Bigg)}^2 \times \dfrac{4}{49}}$

$=\,\,\,$ $\displaystyle \dfrac{2}{5} \times \Large \lim_{u \,\to\, 0} \, \normalsize \dfrac{\sin^2 {\Bigg(\dfrac{7u}{2}\Bigg)}}{{\Bigg(\dfrac{7u}{2}\Bigg)}^2} \times \dfrac{1}{\dfrac{4}{49}}$

$=\,\,\,$ $\displaystyle \dfrac{2}{5} \times \Large \lim_{u \,\to\, 0} \, \normalsize \dfrac{\sin^2 {\Bigg(\dfrac{7u}{2}\Bigg)}}{{\Bigg(\dfrac{7u}{2}\Bigg)}^2} \times \dfrac{49}{4}$

$=\,\,\,$ $\displaystyle \dfrac{2}{5} \times \dfrac{49}{4} \times \Large \lim_{u \,\to\, 0} \, \normalsize \dfrac{\sin^2 {\Bigg(\dfrac{7u}{2}\Bigg)}}{{\Bigg(\dfrac{7u}{2}\Bigg)}^2}$

$=\,\,\,$ $\displaystyle \dfrac{2 \times 49}{5 \times 4} \times \Large \lim_{u \,\to\, 0} \, \normalsize \dfrac{\sin^2 {\Bigg(\dfrac{7u}{2}\Bigg)}}{{\Bigg(\dfrac{7u}{2}\Bigg)}^2}$

$=\,\,\,$ $\require{cancel} \displaystyle \dfrac{\cancel{2} \times 49}{5 \times \cancel{4}} \times \Large \lim_{u \,\to\, 0} \, \normalsize \dfrac{\sin^2 {\Bigg(\dfrac{7u}{2}\Bigg)}}{{\Bigg(\dfrac{7u}{2}\Bigg)}^2}$

$=\,\,\,$ $\require{cancel} \displaystyle \dfrac{49}{5 \times 2} \times \Large \lim_{u \,\to\, 0} \, \normalsize \dfrac{\sin^2 {\Bigg(\dfrac{7u}{2}\Bigg)}}{{\Bigg(\dfrac{7u}{2}\Bigg)}^2}$

$=\,\,\,$ $\require{cancel} \displaystyle \dfrac{49}{10} \times \Large \lim_{u \,\to\, 0} \, \normalsize \dfrac{\sin^2 {\Bigg(\dfrac{7u}{2}\Bigg)}}{{\Bigg(\dfrac{7u}{2}\Bigg)}^2}$

Apply, the quotient rule of exponents to simplify it further.

$=\,\,\,$ $\displaystyle \dfrac{49}{10} \times \Large \lim_{u \,\to\, 0} \, \normalsize {\Bigg(\dfrac{\sin {\Big(\dfrac{7u}{2}\Big)}}{\Big(\dfrac{7u}{2}\Big)} \Bigg)}^2$

$=\,\,\,$ $\displaystyle \dfrac{49}{10} \times {\Bigg(\Large \lim_{u \,\to\, 0} \, \normalsize \dfrac{\sin {\Big(\dfrac{7u}{2}\Big)}}{\Big(\dfrac{7u}{2}\Big)} \Bigg)}^2$

04

Obtaining required result

Try limit of $\dfrac{\sin x}{x}$ rule as $x$ tends to $0$ to get the required result. In order to apply this formula, transform our expression to this form.

If $u \to 0$, then $\dfrac{7}{2} \times u \to \dfrac{7}{2} \times 0$. Therefore, $\dfrac{7u}{2} \to 0$

$=\,\,\,$ $\displaystyle \dfrac{49}{10} \times {\Bigg(\Large \lim_{\huge \frac{7u}{2} \Large \,\to\, 0} \, \normalsize \dfrac{\sin {\Big(\dfrac{7u}{2}\Big)}}{\Big(\dfrac{7u}{2}\Big)} \Bigg)}^2$

$=\,\,\,$ $\displaystyle \dfrac{49}{10} \times {(1)}^2$

$=\,\,\,$ $\displaystyle \dfrac{49}{10} \times 1$

$=\,\,\,$ $\displaystyle \dfrac{49}{10}$

$=\,\,\, 4.9$

It is the required result for this limit problem of the calculus mathematics.



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